Answer to Question #70770 in General Chemistry for tv

Question

Consider the following reaction: 2SO2(g) + O2(g) ->> 2SO3(g)?
if 285.3 mL of SO2 is allowed to react with 158.9 mL of 02(both measured at 315 K and 50 mmHg) what is the limiting reactant and theoretical yield of SO3 in moles? If 187.2 mL of SO3 is collected measured at 315K and 50 mm Hg, what is the percent yield for the reaction?(use ideal gas law)

Answer 1

As one can notice from the reaction equation, two moles of SO2 react with one mole of O2, producing 2 moles of SO3:

n(SO_2 )/2=n(O_2 )/1=n(SO_3 )/2 .
Let’s calculate the number of the moles of gases according to ideal gas law to find out the limiting reactant:

n(SO_2 )=pV/RT=(50 (mmHg)·285.3·10^(-3) L)/(62.36367 (L mmHg K^(-1) mol^(-1) )·315 (K))=7.262·10^(-4) mol,

n(O_2 )=pV/RT=(50 (mmHg)·158.9·10^(-3) L)/(62.36367 (L mmHg K^(-1) mol^(-1) )·315 (K)) = 4.044·10^(-4) mol.

Then, let’s compare the relative quantities of reagents:
(7.262·10^(-4))/2<= 4.044·10^(-4).
As the left term is inferior, it means that SO2 is the limiting reactant.
Considering the next part of the question, we must analogically calculate the number of the moles of SO3:

n(SO_3 )=pV/RT=(50 (mmHg)·187.2·10^(-3) L)/(62.36367 (L mmHg K^(-1) mol^(-1) )·315 (K))=4.765·10^(-4) mol

Theoretical yield of SO3 equals 7.262·10^(-4) mol, as the number of the moles of SO2 and SO3 produced are equal (first equation) and SO2 is the limiting reactant.
Finally, we calculate the experimental yield for the reaction:

%=(4.765·10^(-4) )/(7.262·10^(-4) )·100%=65.6 %

Answer: SO2 is the limiting reactant; experimental yield is 65.6%.